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PROFESSOR: Welcome
to this session

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on separable equations.

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So in this problem, you're asked
in the first question to

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solve the initial value problem,
dydx equals y squared

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with the initial condition
y of zero equals 1.

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In the second part of the
problem, you're asked to find

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the general solution where no
initial condition is imposed.

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So here you need to remember
your method of separation of

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variables to tackle the
first question.

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And then in the second part of
the problem, remember all the

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types of solutions and
conditions that you applied

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the first part to recover
the lost solutions.

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So why don't you take a minute,
pause the video, work

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through questions a and b.

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And then we'll continue together
when I come back.

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Welcome back.

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So in the first part of the
problem, we'll be solving the

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equation dydx equals
y squared.

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So here the method of separation
of variables tells

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us that we should regroup the
variables of the same kind, so

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all the y variables on one side
and dx variable on the

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other side of the equation
and then

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integrate from this point.

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So here for this step, notice
that I divided by y squared,

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which means that we need to
impose the condition y naught

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equal to zero from now on.

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So from this step, we just use
indefinite integrals to

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integrate both sides
of the equation.

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So the left-hand side is the
integral dy over y squared.

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So integral of this gives
us minus 1 over y.

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And the right hand side,
integral dx, is just x.

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Both sides would give us
constant of integrations, but

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we only need one because
this is a first-order

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differential equation.

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And so we group them together
on the right-hand side with

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constant c.

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So from this point, given that
we're interested in variable

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y, we need just to invert
the expression.

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And that gives us partial
answer, which is y of x equals

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minus 1 over x plus c.

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So now we need to use our
initial condition, y of zero

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equals to 1 to determine the
value of c for this particular

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initial value problem.

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So our initial condition
was y of zero equals 1.

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So if we substitute this in the
expression that we just

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obtained, we just have zero plus
c, which then only gives

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us c equal to 1.

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So we end up with the value
for our constant of

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integration, c equals minus 1.

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And so the solution to this
problem is y of x equals 1

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over 1 minus x.

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So if you examine this
expression, you see right away

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that we have a problem for x
equals 1 because at x equals

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1, we have 1/0 which means that,
then, the solution blows

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up and we have a vertical
asymptote.

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So let me draw this here.

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So we're going to have an
asymptote at x equals 1 and a

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solution that passes through
our initial condition, y

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equals 1, going to infinity when
approaching x equals 1.

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But then on the right side of
the value x equals 1, we have

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another part of the solution
that goes to zero

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as x goes to infinity.

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And that diverges to minus
infinity when x approaches 1.

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So by convention, the solutions
of differential

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equations are defined on
one single interval.

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So we need here to realize that
the solution we had is

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basically two parts, the parts
on the left of the asymptote

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and the part on the right
of the asymptote.

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So the solution to our initial
value problems needs to be the

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solution that passes through the
imposed initial condition,

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which was y of zero
equals to 1.

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So it needs to be
this solution.

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So now, if we move on to the
solution of the second part of

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the problem b, we were asked to
find the general solution

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of the problem, which means that
we need to account now

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for all the solutions,
regardless of

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their initial condition.

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So we already answered this
partially during the solution

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of part a where we solved using
indefinite integrals and

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arrived to the solution minus 1
over x plus c, where here we

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had, basically, an undetermined
constant of

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integration.

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So this is one general
solution.

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But remember that we need
to give all the

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solutions of the problem.

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So when we arrived at the
solution, we excluded the

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solution y equals zero, which
was basically a lost solution

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because we had to impose the
condition y not equal to zero.

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So we need when we give the
general solution to this

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differential equation to recover
the lost solution.

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And then we basically have one
kind of solution, minus 1 over

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x plus c, that excludes y equal
to zero and another kind

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of solution that is simply
the zero solution.

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So to summarize, the important
points of this problem is to

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remember the separation of
variables and how to use it

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and the fact that using it
imposes conditions that

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require us to recover lost
solutions at the end of the

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problem if we're asked to
give general solution.

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Another point to remember is
that even a simple ODE can

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lead to relatively complex
behavior which drives the

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presence of this vertical
asymptote that you need to

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then know how to deal with and
determine which part of the

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solutions that you've obtained
is the real solution to the

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initial value problem
that you're given.

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So I hope that you are OK with
this problem and you will use

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these approaches many times for
the rest of the course.

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